8.2 Orthogonal Diagonalization
Theorem 1
The following conditions are equivalent for an (n x n)
matrix P:
1. P is invertible and P-1 = PT
2. The rows of P are orthonormal.
3. The columns of P are orthonormal.
Proof: (1) (2) Given P-1 = PT, PPT = I
Let X1,…,Xn be rows of P. Then XjT is jth col of PT
So (i,j) entry of PPT is Xi•Xj
Since PPT=0, we know that all (i,j) entries where i≠j will be
0, so Xi•Xj=0 (rows are orthog) if i≠j and also Xi•Xj=1 if i=j
(rows are orthonormal) (1/3 equiv is same idea)
Definition
An (n x n) mtx P is called an orthogonal matrix if it satisfies
one (and so all) of the conditions in Theorem 1.
Example
Is the following matrix orthogonal for any angle?
 cos x

 sin x
sin x 

cos x 
Show rows are orthonormal.
Example
If P and Q are orthogonal matrices, show that PQ is also
orthogonal, and that P-1=PT is orthogonal.
Solution: P and Q are invertible, so PQ is also
(PQ)-1 = Q-1P-1=QTPT = (PQ)T so PQ is orthogonal.
Also (P-1)-1 = P = (PT)T = (P-1)T so P-1 is orthogonal.
Definition
An (n x n) matrix A is orthogonally diagonalizable when an
orthogonal matrix P can be found such that P-1AP=PTAP is
diagonal.
(We will see that this only happens with symmetric matrices.)
Theorem 2-Principal Axes Theorem
or Real Spectral Theorem
The following conditions are equivalent for A (n x n):
1. A has an orthonormal set of n eigenvectors. (w/
eigenvalues that we call the spectrum of the matrix.
2. A is orthogonally diagonalizable.
3. A is symmetric.
Proof: (1)(2) P = [X1 … Xn] (n x n)
P is orthog iff {X1,…,Xn} is orthonormal set in n (by thm1)
P-1AP is diag iff {X1,…,Xn} consists of the eigenvectors of A
(from 6.2)
Proof-cont.
(2)(3) If PTAP = D is diagonal where PT=P-1(given from 2),
then A = PDP-1=PDPT
So, AT = (PDPT)T=PTTDTPT=PDTPT=PDPT = A (since D diag)
So A is symmetric.
(3)(2) very tough to prove -- omitted
Definition
Principal Axes: A set of orthonormal eigenvectors of a
symmetric matrix A.
Example
Find an orthogonal matrix P such that P-1AP is diagonal.
3

A  0


0
0
2
2
0 
 We know that A is
2
 orthogonally diagonalizable
since it is symmetric (Thm 2)
5 

1. Find eigenvalues and eigenvectors
2. Make sure you get an orthogonal matrix for P (may need to
use G-S alg)
3. Normalize
Theorem 3
If A is an (n x n) symmetric matrix, then (AX)• Y = X•(AY)
for all columns X and Y in n.
Proof: X • Y = XTY for columns X and Y.
So: (AX) •Y = (AX)TY (by def of dot product)
=XTATY = XTAY (since A is symmetric)
= X • (AY) •
Theorem 4
If A is a symmetric matrix, then eigenvectors of A
corresponding to distinct eigenvalues are orthogonal.
Proof: Let AX = X and AY = Y where  ≠ (so 2 distinct
eigenvalues). Then (using thm 3):
(X•Y) = (X)•Y=(AX)•Y = X•(AY) = X•(Y) = (X•Y)
So: (- ) (X•Y) = 0,
so (X•Y) = 0 since ≠  which tells us that the eigenvectors are
orthogonal. •
Example
3 0 0 


A  0 2 2



0 2 5 

1. Clearly orthogonally diagonalizable since symmetric
Diagonalize:
2. Find eigenvalues
3. Find eigenvectors
4. Eigenvectors of one eigenvalue are orthogonal by thm 4, but if
an eigenvalue has multiplicity > 1, then it also has >1
eigenvector. Its diff’t eigenvalues may not be orthogonal, so you
may need to use G-S alg.
5. Normalize to get orthonormal vectors.
Theorem 5-Triangulation Thm
If A is an (n x n) matrix with real eigenvalues, an orthogonal
matrix P exists such that PTAP is upper triangular.
Proof: much like the proof we omitted earlier.
Corollary
Since a matrix and its diagonalized form are similar, they will
have the same determinant and trace, we can say:
If A is an (n x n) matrix with real eigenvalues 1, 2,…, n (may
not all be distinct), then detA= 1 2... n and
tr A= 1+ 2+…+ n. (since these are the values we get in the
for det and tr of the diagonalized matrix)